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2 years ago

Are oxides bases?

Chemistry

1 answer:

olasank [31]2 years ago

5 0

Answer:

yes

Explanation:

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What happens in nuclear fission reactions?

Dmitry [639]

Answer:

in nuclear fission, an unstable atom splits into 2 or more smaller pieces that are more stable and releases energy in the process. the fission process also releases extra neutrons which can split additional atoms, resulting in a chain reaction that releases a lot of energy

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2 years ago

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there are 1540 megawatts of wind-generated electricity produced globally every year. This amount is equivalent to how many kilow

Reptile [31]

1.54 x 10⁶kW

Explanation:

The problem here is converting from megawatts into kilowatts.

Given 1540megawatts.

The mega- and kilo- watts are prefixes that denotes multiples of units.

1 megawatt = 10⁶watts

1 kilowatt = 10³watts

Now given 1540megawatts power to Kilowatt power;

We see that :

1000kW = 1mW

1540mW x $\frac{1000kW}{1mW}$ = 1.54 x 10⁶kW

learn more:

energy units brainly.com/question/4791744

#learnwithBrainly

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3 years ago

How to find the number of neutrons in an isotope?

Alexeev081 [22]

Chemical properties of elements are primarily determined by the electrons but not by neutrons. Therefore, the isotopes of the same element have similar chemical behavior. (a) The atomic number is 17, so there are 17 protons and 17 electrons. The mass number is 35, so there are 18 (=35-17) neutrons. Hope this helps. :)

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3 years ago

Which reaction has a positive δs°? question 6 options:

Nastasia [14]

B. Because there are 3 molecules in right and 2 molecules in the left, so entropy rises.

3 0

3 years ago

If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under

WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

$n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1$

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

$n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol$

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\ then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L$

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

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3 years ago